Power series representation of $\frac{1}{1-e^z}$ to classify singularity.

Question

I want to classify the singularity of $\frac{1}{1-e^z}$ at $0$.

With a "direct" power series expansion, I got $$\frac{1}{-\sum_{k=1}^\infty \frac{z^k}{k!}}$$

This looks like an essential singularity, but it is not in the standard form for a power series, and the series is in the denominator, so I can't just pull the reciprocal in.

Is there a way to transform this expression into a power series? Or is there a better way to show the singularity is essential?

Answer 1

Let $f(z)=1/(1-e^z)$. There's a simple pole at $0$, because $zf(z)=-1$ at $z=0$, and $zf(z)$ is holomorphic and nonzero in a nbhd of $0$. Indeed, $zf(z)=1/-(1+z/2+z^2/6+\dots)=-1-z/2-5/12z^2+\dots$.